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3 years ago

N the reaction 2Cu2O(s) → 4Cu(s) + O2(g) how does the oxidation number of the copper atoms change

Chemistry

1 answer:

Usimov [2.4K]3 years ago

8 0

Answer: The oxidation number of Cu decreased

Explanation: In Cu2O equation Cu has an oxidation number of

Cu+1 O-2

then in the product side Cu is represented as a single element.

The rule on oxidation number for an individual element is always O

So Cu has 0 oxidation number.

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Which of the following shows the 3 main jobs of a cell?

skad [1K]

C is the answer hope this helps

8 0

3 years ago

aspirin C6H4(CO2)(CO2CH3),can be prepared in the chemistry laboratory by the reactions of salicylic acid, C6H4(CO2H)(OH),with ac

Ket [755]

262mol 1=kg
g=1000
from stoichoimetry

x=102*1000/360
x=102000/360
x=283.33

density =m/v
=283.33/1.082
=262mol

8 0

4 years ago

You have to prepare 100.0 mL of a 0.100 M solution of sodium carbonate. You have a concentrated solution of sodium carbonate tha

Georgia [21]

Answer:

6.9 ml of concentrate

Explanation:

100 ml of .1 M will require .01 moles

from a 1.45 M solution, .01 mole would be

.01 mole / ( 1.45 mole / liter) = 6.9 ml of the concentrate then dilute to 100 ml

4 0

2 years ago

A 16.0 mL sample of a 1.04 M potassium sulfate solution is mixed with 14.3 mL of a 0.880 M barium nitrate solution and this prec

stiks02 [169]

Number of moles in the K2SO4 sample
= (16/1000)*1.04= 0.01664 mol

Number of moles in the Ba(NO3)2 sample
= (14.3/1000*0.880)= 0.01258 mol

Since the reaction is a 1:1 ratio between the two reactants, the limiting reagent is the one containing a smaller number of moles, namely Ba(NO3)2.

The molecular mass of BaSO4 is 137.3+(32.06+4*16.00)=233.4
Therefore the theoretical yield of Barium Sulphate is
233.4*0.01258=2.937 g
Actual yield = 2.60 g (given)
Therefore the percentage yield = 2.60/2.937=88.54%

Answer:
1. the limiting reagent is Barium Nitrate (Ba(NO3)2)
2. the theoretical yield is 2.94 g
3. the percentage yield is 88.5%

I apologize for the mistake previous to this update.

5 0

3 years ago

Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N 2 ( g ) + 3 H 2 ( g )

Kisachek [45]

Answer:

After complete reaction, 0.280 moles of ammonia are produced

Explanation:

Step 1: Data given

Number of moles N2 = 0.140 moles

Number of moles H2 = 0.434 moles

Step 2: The balanced equation

N2(g) + 3H2 (g) ⟶ 2NH3 (g)

Step 3: Calculate the limiting reactant

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

N2 is the limiting reactant. It will completely be consumed (0.140 moles).

H2 is in excess. There will react 3*0.140 = 0.420 moles

There will remain 0.434 - 0.420 = 0.014 moles

Step 4: Calculate moles NH3

For 0.140 moles N2 we'll have 2*0.140 = 0.280 moles NH3

After complete reaction, 0.280 moles of ammonia are produced

5 0

3 years ago